Solutions to Business Mathematics Assignment

# Solutions to Business Mathematics Assignment Solutions to the assignment in Business Mathematics for ND 1 Accountancy ## (1a)If $^{m+n} P_2 = 56$ and $^{m-n} P_2=12$, find m and n Solution: $^{m+n} P_2$ =$\frac{(m+n)!}{(m+n-2)!} =56$ ... (1) $^{m-n} P_2=\frac{(m-n)!}{(m-n-2)!} =12$... (2) From equation 1, we have $\frac{(m+n)!}{(m+n-2)!}$ =$\frac{(m+n)(m+n-1)(m+n-2)!}{(m+n-2)!}$ =56 (m+n)(m+n-1)=56 $(m+n)(m+n-1)=8\times7$ $m+n = 8$...(1) $\frac{(m-n)!}{(m-n-2)!} = \frac{(m-n)(m-n-1)(m-n-2)!}{(m-n-2)!} =12$ $(m-n)(m-n-1)= 12$ $(m-n)(m-n-1)= 4\times 3$ $m-n= 4$...(2) Solving eq(1) and eq(2) by method of elimination: $m+n=8$ $m-n=4$ $2m= 12$ $m=6$ If m=6, then n=8-6 ; n=2 $m=6$, $n=2$ (1b)If $^{n} P_r = 336$ and $^{n} C_r =56$, find n and r Solution: It is well-known that: $^{n} C_r =\frac{^{n} P_r}{r!}$ Then $56 =\frac{336}{r!}$ $r! = \frac{336}{56}$ $r!=6$ $r(r-1)=3\times2$ $r=3$ If we have $^{n} C_3 =56$ $\frac{n!}{(n-3)!3!} =56$ $\frac{n(n-1)(n-2)(n-3)!}{(n-3)!3!} =56$ $\frac{n(n-1)(n-2)}{3\times2} =56$ $n(n-1)(n-2)= 8\times7\times6$ $\Rightarrow$ n= 8 (2)In a survey of 1000 families it is found that 454 use electricity, 502 use gas, 448 use kerosene, 158 use gas and electricity, 160 use gas and kerosene and 134 use electricity and kerosene for cooking. If all of them use at least one of the three, find how many use all the three fuels. Solution: Let us take E for electricity, G for gas, K for kerosene. Now $n(E) = 454$, $n(G) = 502$, $n(K) = 448$. $n( G ∩ E) = 158$ , $n( G ∩ K) = 160$, $n(E ∩ K) = 134$, $n(E ∩ G ∩ K) = ?$ $n(E ∪ G ∪ K) = 1000$ Again $n(E ∪ G ∪ K) = n(E) + n(G) + n(K) – n(E ∩ G)– n(G ∩ K)– n(K ∩ E)+ n(E ∩ G ∩ K)$ $1000 = 454 + 502 + 448 – 158 – 160 –134 + n(E ∩ G ∩ K)$ $1000 = 952 + n(E ∩ G ∩ K)$ $n(E ∩ G ∩ K) = 1000 – 952 = 48$ (3) A student is to answer 8 out of 10 questions on an examination : (i) How many choice has he? (ii) How many if he must answer the first three questions? (iii) How many if he must answer at least four of the first five questions Solution: (i)The 8 questions out of 10 questions may be answered in $^{10} C_8$ $\frac{10!}{(10-8)!8!}=\frac{10\times9\times8!}{2!8!}=5\times9=45ways$ (ii)The first 3 questions are to be answered. So there are remaining 5 (= 8 – 3) questions to be answered out of remaining 7 ( = 10 – 3) questions which may be selected in $^{7} C_5$ways. Now, $^{7} C_5$ = 7.6 = 42 ways. (iii) Here we have the following possible cases : (a) 4 questions from first 5 questions (say, group A), then remaining 4 questions from the balance of 5 questions (say, group B). (b) Again 5 questions from group A, and 3 questions from group B For (a), number of choice is $^{5} C_4$ × $^{5} C_4$= 5 × 5 = 25 For (b) number of ways is $^{5} C_5$×$^{5} C_3$ = 1 × 10 = 10. Hence, Required no. of ways = 25 + 10 = 35. (3)Out of 7 muslims and 15 christians, a committee of 3 muslims and 8 christians is to be formed. In how many ways can this be done if (i) there is no restrictions (ii) 1 particular christian must be included(iii) 2 particular muslims cannot be included on the committee. Solution: (i)$^{7} C_3$ × $^{15} C_8$= 35 × 6435 = 225225ways (ii)$^{7} C_3$ × $^{15-1} C_{8-1}$= $^{7} C_3$ × $^{14} C_7$= 120120ways (iii)$^{7-2} C_3$ × $^{15} C_{8}$= $^{5} C_3$ × $^{15} C_8$= 63450ways A man has 3 friends. In how many ways can be invite one or more of them to dinner? Solution: The number of ways = $^{3} C_1$ + $^{3} C_2$ + $^{3} C_3$ = 3+3+1= 7 ways